Final answer:
The question involves solving a problem based on vector addition and the parallelogram law. By using the given condition that the original resultant force is bisected by the new resultant after halving one of the forces, and applying the law of cosines, we deduce that the 'A' must equal 120 degrees to satisfy the conditions of the problem.
Step-by-step explanation:
We have a situation where two forces of equal magnitudes act on a particle at an angle 'A'. When one force is halved and the resultant force changes, the angle between the unaltered force and the original resultant is bisected by the new resultant force. We can evaluate this geometrically using vector addition. Let the magnitude of each force be F, such that when one force is halved, it becomes F/2. Let R be the original resultant force, and R' be the resultant force after halving one of the forces. By the given condition, the angle between the original force F and the resultant R after one force is halved, is bisected by the new resultant R'. This means that the angle between the force F and R' is A/2, and similarly, the angle between R' and the halved force F/2 is also A/2. From the parallelogram law of vector addition, we know that the diagonals represent the resultant vectors. As one of the diagonals (R') bisects the angle A formed by forces F and F/2, we can set up the law of cosines for two triangles that are formed: one with sides F, F/2, and R', and the other with sides R, F, and R'.
For the triangle with sides F, F/2, and R', we get:
R'^2 = F^2 + (F/2)^2 - (F)(F/2)cos(A/2)
For the triangle with sides R, F, and R', we have:
R'^2 = R^2 + F^2 - 2RFcos(60 degrees)
Since R' is common in both equations and by the condition that the angle is bisected (indicating an isosceles triangle with sides F and R'), we have R = 2(F/2) = F. Hence, R^2 = F^2. Equating the two R'^2 expressions gives a solvable equation for A which turns out to be 120 degrees after simplifying. The given condition is satisfied only when A equals 120 degrees.